AK F.*ing leetcode 流浪计划之多点共面（线）

欢迎关注更多精彩
关注我，学习常用算法与数据结构，一题多解，降维打击。

本期话题，如何判断多点共线，多点共面

如何判断3点共线

采用向量法判断

$向量对于V_1=(x_1,y_1,z_1),V_2=(x_2,y_2,z_2)$

$叉积X=V_1{V}_2=(y_1\ast z_2-z_1\ast y_2,z_1\ast x_2-x_1\ast z_2,x_1\ast y_2-y_1\ast x_2)$

$如向量X是一个0向量，则说明V_1,V_2,共线，对应向量的起终点也共线。$

$对于二维平面叉积是X=(0,0,x_1\ast y_2-y_1\ast x_2),只要判断最后一项是否为0即可$

$给出3点P_1(x_1,y_1,z_1),P_2(x_2,y_2,z_2),P_3(x_3,y_3,z_3)$

$只要判断向量(P_2-P_1),(P_3-P_1)的叉积即可。$

如何判断多点共面

基本思想，先三点（不共线）确定一个平面，然后将第4个点代入平面方程验证。
$平面方程Ax+By+Cy+D=0$
三个不共线的点确定一个平面有2种方法：1.行列式法，2.点法式
$设三点为：P_1(x_1,y_1,z_1),P_2(x_2,y_2,z_2),P_3(x_3,y_3,z_3)$

行列式法

将三点代入方程得到三个方程
通过行列式法求出
$A=\left|\begin{array}{ccc}1& y_1& z_1\\ 1& y_2& z_2\\ 1& y_3& z_3\end{array}\right|$

$B=\left|\begin{array}{ccc}{x}_1& 1& z_1\\ x_2& 1& z_2\\ x_3& 1& z_3\end{array}\right|$

$C=\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$D=-\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|$

$A=y_1(z_2-z_3)+y_2(z_3-z_1)+y_3(z_1-z_2)$

$B=z_1(x_2-x_3)+z_2(x_3-x_1)+z_3(x_1-x_2)$

$C=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$

$D=-x_1(y_2\ast z_3-y_3\ast z_2)-x_2(y_3\ast z_1-y_1\ast z_3)-x_3(y_1\ast z_2-y_2\ast z_1)$

点法式

$P=(P2-P1)X(P3-P1)=(P_x,P_y,P_z)$

$则P为平面的法向量$

$A=P.x,B=P.y,C=P.z$

$将P_1代入方程$

$D=-(A\ast P_{1x}+B\ast P_{1y}+C\ast P_{1z})$

注意点

有可能给出的点都共线，可以先用共线方法判断是否共线，如果没有。
则选择三个不共线的点组成一个平面。
得到方程就可以将剩下的点代入验证。

代码实现

#include 
#include 

using namespace std;

class Point {
public:
    int x, y, z;

    Point(int xx, int yy, int zz) : x(xx), y(yy), z(zz) {}

    Point() {}

    Point operator-(Point &a) const {
        return Point(x - a.x, y - a.y, z - a.z);
    }
};

bool isZero(Point p) {
    return p.x == 0 && p.y == 0 && p.z == 0;
}

Point xmult(Point p1, Point p2) {
    return Point(p1.y * p2.z - p1.z * p2.y, p1.z * p2.x - p1.x * p2.z, p1.x * p2.y - p1.y * p2.x);
}

// 判断多点是否共面，没有重复的点，行列式法
/*
 * 平面方程：Ax+By+Cz+D=0
 * 先选出不共线三点计算出A,B,C，否则如果不存在非共线的点，则都在一平面上。
 * 设有三点P1(x1,y1,z1), P2(x2,y2,z2), P3(x3,y3,z3)
 * A=y1(z2-z3)+y2(z3-z1)+y3(z1-z2)
 * B=z1(x2-x3)+z2(x3-x1)+z3(x1-x2)
 * C=x1(y2-y3)+x2(y3-y1)+x3(y1-y2)
 * D=-x1(y2z3-y3z2)-x2(y3z1-y1z3)-x3(y1z2-y2z1)
 * 将其他点代入平面方程，检查是否为0即可
 */
bool IsInPlane(vector points) {
    int d3 = -1;
    for (int i = 2; i < points.size(); ++i) {
        Point c = xmult(Point(points[1].x - points[0].x, points[1].y - points[0].y, points[1].z - points[0].z),
                        Point(points[i].x - points[0].x, points[i].y - points[0].y, points[i].z - points[0].z));
        if (!isZero(c)) {
            d3 = i;
            break;
        }
    }
    // 所有点都共线
    if (d3 == -1) {
        return true;
    }

    vector planePoints = {points[0], points[1], points[d3]};
    // 计算ABCD
    int A = 0;
    for (int i = 0; i < 3; ++i) {
        A += planePoints[i].y * (planePoints[(i + 1) % 3].z - planePoints[(i + 2) % 3].z);
    }
    int B = 0;
    for (int i = 0; i < 3; ++i) {
        B += planePoints[i].z * (planePoints[(i + 1) % 3].x - planePoints[(i + 2) % 3].x);
    }
    int C = 0;
    for (int i = 0; i < 3; ++i) {
        C += planePoints[i].x * (planePoints[(i + 1) % 3].y - planePoints[(i + 2) % 3].y);
    }

    int D = 0;
    for (int i = 0; i < 3; ++i) {
        D -= planePoints[i].x *
             (planePoints[(i + 1) % 3].y * planePoints[(i + 2) % 3].z -
              planePoints[(i + 2) % 3].y * planePoints[(i + 1) % 3].z);
    }

    // 检查剩余点
    for (int i = 0; i < points.size(); ++i) {
        if (A * points[i].x + B * points[i].y + C * points[i].z + D != 0) {
            return false;
        }
    }

    return true;
}


// 判断多点是否共面，没有重复的点, 法点式
/*
 * 平面方程：Ax+By+Cz+D=0
 * 先选出不共线三点计算出A,B,C，否则如果不存在非共线的点，则都在一平面上。
 * 设有三点P1(x1,y1,z1), P2(x2,y2,z2), P3(x3,y3,z3)
 * P=xmult(P2-P1, P3-P1), 求叉积得到法向
 * A=P.x, B=P.y, C=P.z, D=-P*P1
 * 将其他点代入平面方程，检查是否为0即可
 */
bool IsInPlane2(vector points) {
    int d3 = -1;
    for (int i = 2; i < points.size(); ++i) {
        Point c = xmult(points[1] - points[0], points[i] - points[0]);
        if (!isZero(c)) {
            d3 = i;
            break;
        }
    }
    // 所有点都共线
    if (d3 == -1) {
        return true;
    }

    Point P = xmult(points[1] - points[0], points[d3] - points[0]);
    int D = -(P.x * points[0].x + P.y * points[0].y + P.z * points[0].z);

    // 检查剩余点
    for (int i = 0; i < points.size(); ++i) {
        if (P.x * points[i].x + P.y * points[i].y + P.z * points[i].z + D != 0) {
            return false;
        }
    }

    return true;
}

vector> getTests() {
    vector> ans;
    ans.push_back({{1, 2, 0},
                   {3, 4, 0}});
    ans.push_back({{1, 0, 0},
                   {2, 0, 0},
                   {3, 0, 0},
                   {4, 0, 0}});
    ans.push_back({{1, 2,  0},
                   {3, 4,  0},
                   {4, 5,  0},
                   {6, 7,  0},
                   {6, 10, 0}});
    ans.push_back({{1,   2, 0},
                   {3,   4, 0},
                   {4,   5, 0},
                   {100, 7, 1}});
    ans.push_back({{1,   2, 10},
                   {3,   4, 1},
                   {4,   5, 2},
                   {100, 7, 10}});
    ans.push_back({{1, 0,  0},
                   {0, 1,  0},
                   {0, 0,  1},
                   {2, -1, 0}});
    ans.push_back({{1, 0,  0},
                   {0, 1,  0},
                   {0, 0,  1},
                   {2, -1, 1}});
    return ans;
}

int main(void) {
    auto tests = getTests();

    for (auto v:tests) {
        cout << "p" << endl;
        cout << IsInPlane(v) << endl;
        cout << IsInPlane2(v) << endl;
    }
    return 0;
}

投影法

取面上一点与目标点形成向量，求向量在法向上的投影是否为0。

练习题目

https://leetcode-cn.com/problems/max-points-on-a-line/

https://leetcode-cn.com/problems/check-if-it-is-a-straight-line/